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Find the Taylor series about the given point for each of the functions. In each case find the radius of convergence. f(x)=sinx2f(x)=\sin x^{2} about x0=0x_{0}=0

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Solution\textbf{Solution}: Let us consider an analytic function ff defined by

f(x)=sinx2.f(x)=\sin x^2.

We will now find the power series expansion of the function ff about the point x0=0.\>x_0=0. First consider that y=x2\> y=x^2. Then our reformed function is f(y)=siny.\> f(y) =\sin y. And also note that x02=y0=0.\> x_0^2=y_0=0. Now notice that

f(y)=n=0f(n)(y0)n!(yy0)n,\begin{align}f(y)=\sum_{n=0}^{\infty} \frac{f^{(n)}(y_0)}{n!}(y-y_0)^n,\end{align}

for all yy in the interval of convergence. Recall that, f(y)=siny.\> f(y)=\sin y. Now notice that

f(1)(y)=cosyf(2)(y)=sinyf(3)(y)=cosyf(4)(y)=siny\begin{align*} & f^{(1)}(y)=\cos y \\ & f^{(2)}(y)=- \sin y \\ & f^{(3)}(y)=-\cos y \\ & f^{(4)}(y)=\sin y \\ & \cdots \cdots \end{align*}

Then notice that, at y0=0y_0=0 we have

f(2n)(0)=0 and f(2n+1)(0)=(1)n.f^{(2n)}(0)=0 \>\>\> \text{ and }\>\>\> f^{(2n+1)}(0)=(-1)^{n}.

Then from (1)(1) we have

f(y)=n=0(1)n(2n+1)!y2n+1.f(y)=\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} y^{2n+1}.

Then notice that

f(x)=n=0(1)n(2n+1)!(x2)2n+1.f(x)=\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} (x^2)^{2n+1}.

Thus the Taylor series expansion of f(x)=sinx2f(x)=\sin x^2 is given by

sinx2=n=0(1)n(2n+1)!x2n+1,\sin x^2= \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} x^{2n+1},

for all xx in the interval of convergence.

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